Problem: The value of $\sqrt{31}$ lies between which two consecutive integers ? Integers that appear in order when counting, for example 2 and 3.
Consider the perfect squares near $31$ . [ What are perfect squares? Perfect squares are integers which can be obtained by squaring an integer. The first 13 perfect squares are: $ 1,4,9,16,25,36,49,64,81,100,121,144,169$ $25$ is the nearest perfect square less than $31$ $36$ is the nearest perfect square more than $31$ So, we know $25 < 31 < 36$ So, $\sqrt{25} < \sqrt{31} < \sqrt{36}$ So $\sqrt{31}$ is between $5$ and $6$.